Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 3e3 + 10, M = 1e3 + 20, mod = 2017, inf = 2e9;
int b[N][N], m, n, a[N][N];
vector<int> ans1, ans2, ans3, ans4;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) scanf("%d", &a[i][j]), b[i][j] = a[i][j];
  }
  int fi = 0;
  for (int i = 1; i <= n; ++i) {
    int sum = inf;
    for (int j = 1; j <= m; ++j) {
      sum = min(sum, a[i][j]);
    }
    if (sum) {
      for (int j = 1; j <= sum; ++j) ans1.push_back(i);
      for (int j = 1; j <= m; ++j) a[i][j] -= sum;
    }
  }
  for (int i = 1; i <= m; ++i) {
    int sum = inf;
    for (int j = 1; j <= n; ++j) {
      sum = min(sum, a[j][i]);
    }
    if (sum) {
      for (int j = 1; j <= sum; ++j) ans2.push_back(i);
      for (int j = 1; j <= n; ++j) a[j][i] -= sum;
    }
  }
  fi = ans1.size() + ans2.size();
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (a[i][j]) {
        puts("-1");
        return 0;
      }
    }
  }
  for (int i = 1; i <= m; ++i) {
    int sum = inf;
    for (int j = 1; j <= n; ++j) {
      sum = min(sum, b[j][i]);
    }
    if (sum) {
      for (int j = 1; j <= sum; ++j) ans4.push_back(i);
      for (int j = 1; j <= n; ++j) b[j][i] -= sum;
    }
  }
  for (int i = 1; i <= n; ++i) {
    int sum = inf;
    for (int j = 1; j <= m; ++j) {
      sum = min(sum, b[i][j]);
    }
    if (sum) {
      for (int j = 1; j <= sum; ++j) ans3.push_back(i);
      for (int j = 1; j <= m; ++j) b[i][j] -= sum;
    }
  }
  int se = ans3.size() + ans4.size();
  if (fi <= se) {
    cout << ans1.size() + ans2.size() << endl;
    for (int i = 0; i < ans1.size(); ++i) {
      cout << "row"
           << " " << ans1[i] << endl;
    }
    for (int i = 0; i < ans2.size(); ++i) {
      cout << "col"
           << " " << ans2[i] << endl;
    }
  } else {
    cout << ans3.size() + ans4.size() << endl;
    for (int i = 0; i < ans3.size(); ++i) {
      cout << "row"
           << " " << ans3[i] << endl;
    }
    for (int i = 0; i < ans4.size(); ++i) {
      cout << "col"
           << " " << ans4[i] << endl;
    }
  }
  return 0;
}